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  • how to run/compile java code from JTextArea at Runtime? ----urgent!!! college project

    - by Lokesh Kumar
    I have a JInternalFrame painted with a BufferedImage and contained in the JDesktopPane of a JFrame.I also have a JTextArea where i want to write some java code (function) that takes the current JInternalFrame's painted BufferedImage as an input and after doing some manipulation on this input it returns another manipulated BufferedImage that paints the JInternalFrame with new manipulated Image again!!. Manipulation java code of JTextArea:- public BufferedImage customOperation(BufferedImage CurrentInputImg) { Color colOld; Color colNew; BufferedImage manipulated=new BufferedImage(CurrentInputImg.getWidth(),CurrentInputImg.getHeight(),BufferedImage.TYPE_INT_ARGB); //make all Red pixels of current image black for(int i=0;i< CurrentInputImg.getWidth();i++) { for(int j=0;j< CurrentInputImg.getHeight(),j++) { colOld=new Color(CurrentInputImg.getRGB(i,j)); colNew=new Color(0,colOld.getGreen(),colOld.getBlue(),colOld.getAlpha()); manipulated.setRGB(i,j,colNew.getRGB()); } } return manipulated; } so,how can i run/compile this JTextArea java code at runtime and get a new manipulated image for painting on JInternalFrame???????   Here is my Main class: (This class is not actual one but i have created it for u for basic interfacing containing JTextArea,JInternalFrame,Apply Button) import java.awt.*; import java.awt.event.*; import javax.swing.*; import javax.swing.event.*; import javax.swing.JInternalFrame; import javax.swing.JDesktopPane; import java.awt.image.*; import javax.imageio.*; import java.io.*; import java.io.File; import java.util.*; class MyCustomOperationSystem extends JFrame **{** public JInternalFrame ImageFrame; public BufferedImage CurrenFrameImage; public MyCustomOperationSystem() **{** setTitle("My Custom Image Operations"); setSize((int)Toolkit.getDefaultToolkit().getScreenSize().getWidth(),(int)Toolkit.getDefaultToolkit().getScreenSize().getHeight()); JDesktopPane desktop=new JDesktopPane(); desktop.setPreferredSize(new Dimension((int)Toolkit.getDefaultToolkit().getScreenSize().getWidth(),(int)Toolkit.getDefaultToolkit().getScreenSize().getHeight())); try{ CurrenFrameImage=ImageIO.read(new File("c:/Lokesh.png")); }catch(Exception exp) { System.out.println("Error in Loading Image"); } ImageFrame=new JInternalFrame("Image Frame",true,true,false,true); ImageFrame.setMinimumSize(new Dimension(CurrenFrameImage.getWidth()+10,CurrenFrameImage.getHeight()+10)); ImageFrame.getContentPane().add(CreateImagePanel()); ImageFrame.setLayer(1); ImageFrame.setLocation(100,100); ImageFrame.setVisible(true); desktop.setOpaque(true); desktop.setBackground(Color.darkGray); desktop.add(ImageFrame); this.getContentPane().setLayout(new BorderLayout()); this.getContentPane().add("Center",desktop); this.getContentPane().add("South",ControlPanel()); pack(); setVisible(true); **}** public JPanel CreateImagePanel(){ JPanel tempPanel=new JPanel(){ public void paintComponent(Graphics g) { g.drawImage(CurrenFrameImage,0,0,this); } }; tempPanel.setPreferredSize(new Dimension(CurrenFrameImage.getWidth(),CurrenFrameImage.getHeight())); return tempPanel; } public JPanel ControlPanel(){ JPanel controlPan=new JPanel(new FlowLayout(FlowLayout.LEFT)); JButton customOP=new JButton("Custom Operation"); customOP.addActionListener(new ActionListener(){ public void actionPerformed(ActionEvent evnt){ JFrame CodeFrame=new JFrame("Write your Code Here"); JTextArea codeArea=new JTextArea("Your Java Code Here",100,70); JScrollPane codeScrollPan=new JScrollPane(codeArea,ScrollPaneConstants.VERTICAL_SCROLLBAR_ALWAYS, ScrollPaneConstants.HORIZONTAL_SCROLLBAR_ALWAYS); CodeFrame.add(codeScrollPan); CodeFrame.setVisible(true); } }); JButton Apply=new JButton("Apply Code"); Apply.addActionListener(new ActionListener(){ public void actionPerformed(ActionEvent event){ // What should I do!!! Here!!!!!!!!!!!!!!! } }); controlPan.add(customOP); controlPan.add(Apply); return controlPan; } public static void main(String s[]) { new MyCustomOperationSystem(); } } Note: in above class JInternalFrame (ImageFrame) is not visible even i have declared it visible. so, ImageFrame is not visible while compiling and running above class. U have to identify this problem before running it.

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  • How to judge the relative efficiency of algorithms given runtimes as functions of 'n'?

    - by Lopa
    Consider two algorithms A and B which solve the same problem, and have time complexities (in terms of the number of elementary operations they perform) given respectively by a(n) = 9n+6 b(n) = 2(n^2)+1 (i) Which algorithm is the best asymptotically? (ii) Which is the best for small input sizes n, and for what values of n is this the case? (You may assume where necessary that n0.) i think its 9n+6. guys could you please help me with whether its right or wrong?? and whats the answer for part b. what exactly do they want?

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  • Entry lvl. COBOL Control Breaks

    - by Kyle Benzle
    I'm working in COBOL with a double control break to print a hospital record. The input is one record per line, with, hospital info first, then patient info. There are multiple records per hospital, and multiple services per patient. The idea is, using a double control break, to print one hospital name, then all the patients from that hospital. Then print the patient name just once for all services, like the below. I'm having trouble with my output, and am hoping someone can help me get it in order. I am using AccuCobol to compile experts-exchange does not allow .cob and .dat so the extentions were changed to .txt The files are: the .cob lab5b.cob the input / output: lab5bin.dat, lab5bout.dat The assignment: http://www.cse.ohio-state.edu/~sgomori/314/lab5.html Hospital Number: 001 Hospital Name: Mount Carmel 00001 Griese, Brian Ear Infection 08/24/1999 300.00 Diaper Rash 09/05/1999 25.00 Frontal Labotomy 09/25/1999 25,000.00 Rear Labotomy 09/26/1999 25,000.00 Central Labotomy 09/28/1999 24,999.99 The total amount owed for this patient is: $.......... (End of Hospital) The total amount owed for this hospital is: $......... enter code here IDENTIFICATION DIVISION. PROGRAM-ID. LAB5B. ENVIRONMENT DIVISION. INPUT-OUTPUT SECTION. FILE-CONTROL. SELECT FILE-IN ASSIGN TO 'lab5bin.dat' ORGANIZATION IS LINE SEQUENTIAL. SELECT FILE-OUT ASSIGN TO 'lab5bout.dat' ORGANIZATION IS LINE SEQUENTIAL. DATA DIVISION. FILE SECTION. FD FILE-IN. 01 HOSPITAL-RECORD-IN. 05 HOSPITAL-NUMBER-IN PIC 999. 05 HOSPITAL-NAME-IN PIC X(20). 05 PATIENT-NUMBER-IN PIC 99999. 05 PATIENT-NAME-IN PIC X(20). 05 SERVICE-IN PIC X(30). 05 DATE-IN PIC 9(8). 05 OWED-IN PIC 9(7)V99. FD FILE-OUT. 01 REPORT-REC-OUT PIC X(100). WORKING-STORAGE SECTION. 01 WS-WORK-AREAS. 05 WS-HOLD-HOSPITAL-NUM PIC 999 VALUE ZEROS. 05 WS-HOLD-PATIENT-NUM PIC 99999 VALUE ZEROS. 05 ARE-THERE-MORE-RECORDS PIC XXX VALUE 'YES'. 88 MORE-RECORDS VALUE 'YES'. 88 NO-MORE-RECORDS VALUE 'NO '. 05 FIRST-RECORD PIC XXX VALUE 'YES'. 05 WS-PATIENT-TOTAL PIC 9(9)V99 VALUE ZEROS. 05 WS-HOSPITAL-TOTAL PIC 9(9)V99 VALUE ZEROS. 05 WS-PAGE-CTR PIC 99 VALUE ZEROS. 01 WS-DATE. 05 WS-YR PIC 9999. 05 WS-MO PIC 99. 05 WS-DAY PIC 99. 01 HL-HEADING1. 05 PIC X(49) VALUE SPACES. 05 PIC X(14) VALUE 'OHIO INSURANCE'. 05 PIC X(7) VALUE SPACES. 05 HL-PAGE PIC Z9. 05 PIC X(14) VALUE SPACES. 05 HL-DATE. 10 HL-MO PIC 99. 10 PIC X VALUE '/'. 10 HL-DAY PIC 99. 10 PIC X VALUE '/'. 10 HL-YR PIC X VALUE '/'. 01 HL-HEADING2. 05 PIC XXXXXXXXXX VALUE 'HOSPITAL: '. 05 HL-HOSPITAL PIC 999. 01 HL-HEADING3. 05 PIC X(7) VALUE "Patient". 05 PIC X(3) VALUE SPACES. 05 PIC X(7) VALUE "Patient". 05 PIC X(39) VALUE SPACES. 05 PIC X(7) VALUE "Date of". 05 PIC X(3) VALUE SPACES. 05 PIC X(6) VALUE "Amount". 01 HL-HEADING4. 05 PIC X(6) VALUE "Number". 05 PIC X(4) VALUE SPACES. 05 PIC X(4) VALUE "Name". 05 PIC X(18) VALUE SPACES. 05 PIC X(10) VALUE "Service". 05 PIC X(14) VALUE SPACES. 05 PIC X(8) VALUE "Service". 05 PIC X(2) VALUE SPACES. 05 PIC X(5) VALUE "Owed". 01 DL-PATIENT-LINE. 05 PIC X(28) VALUE SPACES. 05 DL-PATIENT-NUMBER PIC XXXXX. 05 PIC X(21) VALUE SPACES. 05 DL-PATIENT-TOTAL PIC $$$,$$$,$$9.99. 01 DL-HOSPITAL-LINE. 05 PIC X(47) VALUE SPACES. 05 PIC X(16) VALUE 'HOSPITAL TOTAL: '. 05 DL-HOSPITAL-TOTAL PIC $$$,$$$,$$9.99. PROCEDURE DIVISION. 100-MAIN-MODULE. PERFORM 600-INITIALIZATION-RTN PERFORM UNTIL NO-MORE-RECORDS READ FILE-IN AT END MOVE 'NO ' TO ARE-THERE-MORE-RECORDS NOT AT END PERFORM 200-DETAIL-RTN END-READ END-PERFORM PERFORM 400-HOSPITAL-BREAK PERFORM 700-END-OF-JOB-RTN STOP RUN. 200-DETAIL-RTN. EVALUATE TRUE WHEN FIRST-RECORD = 'YES' MOVE PATIENT-NUMBER-IN TO WS-HOLD-PATIENT-NUM MOVE HOSPITAL-NUMBER-IN TO WS-HOLD-HOSPITAL-NUM PERFORM 500-HEADING-RTN MOVE 'NO ' TO FIRST-RECORD WHEN HOSPITAL-NUMBER-IN NOT = WS-HOLD-HOSPITAL-NUM PERFORM 400-HOSPITAL-BREAK WHEN PATIENT-NUMBER-IN NOT = WS-HOLD-PATIENT-NUM PERFORM 300-PATIENT-BREAK END-EVALUATE ADD OWED-IN TO WS-PATIENT-TOTAL. 300-PATIENT-BREAK. MOVE WS-PATIENT-TOTAL TO DL-PATIENT-TOTAL MOVE WS-HOLD-PATIENT-NUM TO DL-PATIENT-NUMBER WRITE REPORT-REC-OUT FROM DL-PATIENT-LINE AFTER ADVANCING 2 LINES ADD WS-PATIENT-TOTAL TO WS-HOSPITAL-TOTAL IF MORE-RECORDS MOVE ZEROS TO WS-PATIENT-TOTAL MOVE PATIENT-NUMBER-IN TO WS-HOLD-PATIENT-NUM END-IF. 400-HOSPITAL-BREAK. PERFORM 300-PATIENT-BREAK MOVE WS-HOSPITAL-TOTAL TO DL-HOSPITAL-TOTAL WRITE REPORT-REC-OUT FROM DL-HOSPITAL-LINE AFTER ADVANCING 2 LINES IF MORE-RECORDS MOVE ZEROS TO WS-HOSPITAL-TOTAL MOVE HOSPITAL-NUMBER-IN TO WS-HOLD-HOSPITAL-NUM PERFORM 500-HEADING-RTN END-IF. 500-HEADING-RTN. ADD 1 TO WS-PAGE-CTR MOVE WS-PAGE-CTR TO HL-PAGE MOVE WS-HOLD-HOSPITAL-NUM TO HL-HOSPITAL WRITE REPORT-REC-OUT FROM HL-HEADING1 AFTER ADVANCING PAGE WRITE REPORT-REC-OUT FROM HL-HEADING2 AFTER ADVANCING 2 LINES. WRITE REPORT-REC-OUT FROM HL-HEADING3 AFTER ADVANCING 2 LINES. 600-INITIALIZATION-RTN. OPEN INPUT FILE-IN OUTPUT FILE-OUT *159 ACCEPT WS-DATE FROM DATE YYYYMMDD MOVE WS-YR TO HL-YR MOVE WS-MO TO HL-MO MOVE WS-DAY TO HL-DAY. 700-END-OF-JOB-RTN. CLOSE FILE-IN FILE-OUT.

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  • creating child process

    - by rishabh
    Write a program to creates two childs by using send –pipe and receive –pipe primitives.Each child Process reads from a different serial line and sends the character read back to parent process through a pie. The parent process outputs all characters received. A child terminates when exclamation point character is received.The parent process terminates after both children have terminate.

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  • How to get predecessor and successors from an adjacency matrix

    - by NickTFried
    Hi I am am trying to complete an assignment, where it is ok to consult the online community. I have to create a graph class that ultimately can do Breadth First Search and Depth First Search. I have been able to implement those algorithms successfully however another requirement is to be able to get the successors and predecessors and detect if two vertices are either predecessors or successors for each other. I'm having trouble thinking of a way to do this. I will post my code below, if anyone has any suggestions it would be greatly appreciated. import java.util.ArrayList; import java.util.Iterator; import java.util.LinkedList; import java.util.Queue; import java.util.Stack; public class Graph<T> { public Vertex<T> root; public ArrayList<Vertex<T>> vertices=new ArrayList<Vertex<T>>(); public int[][] adjMatrix; int size; private ArrayList<Vertex<T>> dfsArrList; private ArrayList<Vertex<T>> bfsArrList; public void setRootVertex(Vertex<T> n) { this.root=n; } public Vertex<T> getRootVertex() { return this.root; } public void addVertex(Vertex<T> n) { vertices.add(n); } public void removeVertex(int loc){ vertices.remove(loc); } public void addEdge(Vertex<T> start,Vertex<T> end) { if(adjMatrix==null) { size=vertices.size(); adjMatrix=new int[size][size]; } int startIndex=vertices.indexOf(start); int endIndex=vertices.indexOf(end); adjMatrix[startIndex][endIndex]=1; adjMatrix[endIndex][startIndex]=1; } public void removeEdge(Vertex<T> v1, Vertex<T> v2){ int startIndex=vertices.indexOf(v1); int endIndex=vertices.indexOf(v2); adjMatrix[startIndex][endIndex]=1; adjMatrix[endIndex][startIndex]=1; } public int countVertices(){ int ver = vertices.size(); return ver; } /* public boolean isPredecessor( Vertex<T> a, Vertex<T> b){ for() return true; }*/ /* public boolean isSuccessor( Vertex<T> a, Vertex<T> b){ for() return true; }*/ public void getSuccessors(Vertex<T> v1){ } public void getPredessors(Vertex<T> v1){ } private Vertex<T> getUnvisitedChildNode(Vertex<T> n) { int index=vertices.indexOf(n); int j=0; while(j<size) { if(adjMatrix[index][j]==1 && vertices.get(j).visited==false) { return vertices.get(j); } j++; } return null; } public Iterator<Vertex<T>> bfs() { Queue<Vertex<T>> q=new LinkedList<Vertex<T>>(); q.add(this.root); printVertex(this.root); root.visited=true; while(!q.isEmpty()) { Vertex<T> n=q.remove(); Vertex<T> child=null; while((child=getUnvisitedChildNode(n))!=null) { child.visited=true; bfsArrList.add(child); q.add(child); } } clearVertices(); return bfsArrList.iterator(); } public Iterator<Vertex<T>> dfs() { Stack<Vertex<T>> s=new Stack<Vertex<T>>(); s.push(this.root); root.visited=true; printVertex(root); while(!s.isEmpty()) { Vertex<T> n=s.peek(); Vertex<T> child=getUnvisitedChildNode(n); if(child!=null) { child.visited=true; dfsArrList.add(child); s.push(child); } else { s.pop(); } } clearVertices(); return dfsArrList.iterator(); } private void clearVertices() { int i=0; while(i<size) { Vertex<T> n=vertices.get(i); n.visited=false; i++; } } private void printVertex(Vertex<T> n) { System.out.print(n.label+" "); } }

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  • CSS : overflow : auto will not work under FireFox 3.6.2

    - by Michael Mao
    Hello everyone: This is a CSS related question, I got one good answer from my previous question, which suggested to use some CSS code like overflow:auto together with a fixed height container. And here is my actual implementation : on uni server Please follow the instructions on screen and buy more than 4 kinds of tickets. If you are using IE8, Opera, Safari, Chrome, you would notice that the lower right corner of the page now has a vertical scroll bar, which scrolls the content inside it and prevent it from overflowing. That's what I want to have in this section. Now the problem is, this would not do in FireFox 3.6.2. Am I doing something not compliant to the CSS standard or FireFox has its own way of overflow control? You can inspect the elements on screen, and all controlling functions are done in one javascript using jQuery. All CSS code are kept in a separated file as well. According to the professor, FireFox would be the target browser, although the version was set to 2.0...

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  • Regular expression for a phone number

    - by Zerobu
    Hello, I would like a regular expression in this format. It Must match one of the following formats: * (###)###-#### * ###-###-#### * ###.###.#### * ########## Strip all whitespace. Make sure it's a valid phone number, then (if necessary) translate it to the first format listed above.

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  • An algorithm for pavement usage calculation

    - by student
    Given an area of specific size I need to find out how many pavement stones to use to completely pave the area. Suppose that I have an empty floor of 100 metre squares and stones with 20x10 cm and 30x10 cm sizes. I must pave the area with minimum usage of stones of both sizes. Anyone knows of an algorithm that calculates this? (Sorry if my English is bad) C# is preferred.

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  • templates of functions

    - by anotr67
    I'm told to create template of function , that will take 4 arguments : pointer reference pointer to array pointer to function How to perform this task ? I was trying : template<typename TYPE> TYPE biggest(TYPE *L, TYPE $M, TYPE *K[], TYPE *O()) { } but it is wrong.

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  • How web browser works ?

    - by Anil Namde
    I have tired to find good documentation of browsers using google but failed to get what i am looking for. Can some one guide me to location where i can actually see how browser functions. The whole purpose of the exercise is to get answers for following queries and more like these.... How images, css and js files are downloaded How js is executed How Ajax request is executed and many more like these..... Thanks all,

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  • Detecting what changed in an HTML Textfield

    - by teehoo
    For a major school project I am implementing a real-time collaborative editor. For a little background, basically what this means is that two(or more) users can type into a document at the same time, and their changes are automatically propagated to one another (similar to Etherpad). Now my problem is as follows: I want to be able to detect what changes a user carried out onto an HTML textfield. They could: Insert a character Delete a character Paste a string of characters Cut a string of characters I want to be able to detect which of these changes happened and then notify other clients similar to "insert character 'c' at position 2" etc. Anyway I was hoping to get some advice on how I would go about implementing the detection of these changes? My first attempt was to consider the carot position before and after a change occurred, but this failed miserably. For my second attempt I was thinking about doing a diff on the entire contents of the textfields old and new value. Am I missing anything obvious with this solution? Is there something simpler?

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  • Thread scheduling C

    - by MRP
    include <pthread.h> include <stdio.h> include <stdlib.h> #define NUM_THREADS 4 #define TCOUNT 5 #define COUNT_LIMIT 13 int done = 0; int count = 0; int thread_ids[4] = {0,1,2,3}; int thread_runtime[4] = {0,5,4,1}; pthread_mutex_t count_mutex; pthread_cond_t count_threshold_cv; void *inc_count(void *t) { int i; long my_id = (long)t; long run_time = thread_runtime[my_id]; if (my_id==2 && done ==0) { for(i=0; i< 5 ; i++) { if( i==4 ){done =1;} pthread_mutex_lock(&count_mutex); count++; if (count == COUNT_LIMIT) { pthread_cond_signal(&count_threshold_cv); printf("inc_count(): thread %ld, count = %d Threshold reached.\n", my_id, count); } printf("inc_count(): thread %ld, count = %d, unlocking mutex\n", my_id, count); pthread_mutex_unlock(&count_mutex); } } if (my_id==3 && done==1) { for(i=0; i< 4 ; i++) { if(i == 3 ){ done = 2;} pthread_mutex_lock(&count_mutex); count++; if (count == COUNT_LIMIT) { pthread_cond_signal(&count_threshold_cv); printf("inc_count(): thread %ld, count = %d Threshold reached.\n", my_id, count); } printf("inc_count(): thread %ld, count = %d, unlocking mutex\n", my_id, count); pthread_mutex_unlock(&count_mutex); } } if (my_id==4&& done == 2) { for(i=0; i< 8 ; i++) { pthread_mutex_lock(&count_mutex); count++; if (count == COUNT_LIMIT) { pthread_cond_signal(&count_threshold_cv); printf("inc_count(): thread %ld, count = %d Threshold reached.\n",my_id, count); } printf("inc_count(): thread %ld, count = %d, unlocking mutex\n", my_id, count); pthread_mutex_unlock(&count_mutex); } } pthread_exit(NULL); } void *watch_count(void *t) { long my_id = (long)t; printf("Starting watch_count(): thread %ld\n", my_id); pthread_mutex_lock(&count_mutex); if (count<COUNT_LIMIT) { pthread_cond_wait(&count_threshold_cv, &count_mutex); printf("watch_count(): thread %ld Condition signal received.\n", my_id); count += 125; printf("watch_count(): thread %ld count now = %d.\n", my_id, count); } pthread_mutex_unlock(&count_mutex); pthread_exit(NULL); } int main (int argc, char *argv[]) { int i, rc; long t1=1, t2=2, t3=3, t4=4; pthread_t threads[4]; pthread_attr_t attr; pthread_mutex_init(&count_mutex, NULL); pthread_cond_init (&count_threshold_cv, NULL); pthread_attr_init(&attr); pthread_attr_setdetachstate(&attr,PTHREAD_CREATE_JOINABLE); pthread_create(&threads[0], &attr, watch_count, (void *)t1); pthread_create(&threads[1], &attr, inc_count, (void *)t2); pthread_create(&threads[2], &attr, inc_count, (void *)t3); pthread_create(&threads[3], &attr, inc_count, (void *)t4); for (i=0; i<NUM_THREADS; i++) { pthread_join(threads[i], NULL); } printf ("Main(): Waited on %d threads. Done.\n", NUM_THREADS); pthread_attr_destroy(&attr); pthread_mutex_destroy(&count_mutex); pthread_cond_destroy(&count_threshold_cv); pthread_exit(NULL); } so this code creates 4 threads. thread 1 keeps track of the count value while the other 3 increment the count value. the run time is the number of times the thread will increment the count value. I have a done value that allows the first thread to increment the count value first until its run time is up.. so its like a First Come First Serve. my question is, is there a better way of implementing this? I have read about SCHED_FIFO or SCHED_RR.. I guess I dont know how to implement them into this code or if it can be.

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  • Error codes for C++

    - by billy
    #include <iostream> #include <iomanip> using namespace std; //Global constant variable declaration const int MaxRows = 8, MaxCols = 10, SEED = 10325; //Functions Declaration void PrintNameHeader(ostream& out); void Fill2DArray(double ary[][MaxCols]); void Print2DArray(const double ary[][MaxCols]); double GetTotal(const double ary[][MaxCols]); double GetAverage(const double ary[][MaxCols]); double GetRowTotal(const double ary[][MaxCols], int theRow); double GetColumnTotal(const double ary[][MaxCols], int theRow); double GetHighestInRow(const double ary[][MaxCols], int theRow); double GetLowestInRow(const double ary[][MaxCols], int theRow); double GetHighestInCol(const double ary[][MaxCols], int theCol); double GetLowestInCol(const double ary[][MaxCols], int theCol); double GetHighest(const double ary[][MaxCols], int& theRow, int& theCol); double GetLowest(const double ary[][MaxCols], int& theRow, int& theCol); int main() { int theRow; int theCol; PrintNameHeader(cout); cout << fixed << showpoint << setprecision(1); srand(static_cast<unsigned int>(SEED)); double ary[MaxRows][MaxCols]; cout << "The seed value for random number generator is: " << SEED << endl; cout << endl; Fill2DArray(ary); Print2DArray(ary); cout << " The Total for all the elements in this array is: " << setw(7) << GetTotal(ary) << endl; cout << "The Average of all the elements in this array is: " << setw(7) << GetAverage(ary) << endl; cout << endl; cout << "The sum of each row is:" << endl; for(int index = 0; index < MaxRows; index++) { cout << "Row " << (index + 1) << ": " << GetRowTotal(ary, theRow) << endl; } cout << "The highest and lowest of each row is: " << endl; for(int index = 0; index < MaxCols; index++) { cout << "Row " << (index + 1) << ": " << GetHighestInRow(ary, theRow) << " " << GetLowestInRow(ary, theRow) << endl; } cout << "The highest and lowest of each column is: " << endl; for(int index = 0; index < MaxCols; index++) { cout << "Col " << (index + 1) << ": " << GetHighestInCol(ary, theRow) << " " << GetLowestInCol(ary, theRow) << endl; } cout << "The highest value in all the elements in this array is: " << endl; cout << GetHighest(ary, theRow, theCol) << "[" << theRow << "]" << "[" << theCol << "]" << endl; cout << "The lowest value in all the elements in this array is: " << endl; cout << GetLowest(ary, theRow, theCol) << "[" << theRow << "]" << "[" << theCol << "]" << endl; return 0; } //Define Functions void PrintNameHeader(ostream& out) { out << "*******************************" << endl; out << "* *" << endl; out << "* C.S M10A Spring 2010 *" << endl; out << "* Programming Assignment 10 *" << endl; out << "* Due Date: Thurs. Mar. 25 *" << endl; out << "*******************************" << endl; out << endl; } void Fill2DArray(double ary[][MaxCols]) { for(int index1 = 0; index1 < MaxRows; index1++) { for(int index2= 0; index2 < MaxCols; index2++) { ary[index1][index2] = (rand()%1000)/10; } } } void Print2DArray(const double ary[][MaxCols]) { cout << " Column "; for(int index = 0; index < MaxCols; index++) { int column = index + 1; cout << " " << column << " "; } cout << endl; cout << " "; for(int index = 0; index < MaxCols; index++) { int column = index +1; cout << "----- "; } cout << endl; for(int index1 = 0; index1 < MaxRows; index1++) { cout << "Row " << (index1 + 1) << ":"; for(int index2= 0; index2 < MaxCols; index2++) { cout << setw(6) << ary[index1][index2]; } } } double GetTotal(const double ary[][MaxCols]) { double total = 0; for(int theRow = 0; theRow < MaxRows; theRow++) { total = total + GetRowTotal(ary, theRow); } return total; } double GetAverage(const double ary[][MaxCols]) { double total = 0, average = 0; total = GetTotal(ary); average = total / (MaxRows * MaxCols); return average; } double GetRowTotal(const double ary[][MaxCols], int theRow) { double sum = 0; for(int index = 0; index < MaxCols; index++) { sum = sum + ary[theRow][index]; } return sum; } double GetColumTotal(const double ary[][MaxCols], int theCol) { double sum = 0; for(int index = 0; index < theCol; index++) { sum = sum + ary[index][theCol]; } return sum; } double GetHighestInRow(const double ary[][MaxCols], int theRow) { double highest = 0; for(int index = 0; index < MaxCols; index++) { if(ary[theRow][index] > highest) highest = ary[theRow][index]; } return highest; } double GetLowestInRow(const double ary[][MaxCols], int theRow) { double lowest = 0; for(int index = 0; index < MaxCols; index++) { if(ary[theRow][index] < lowest) lowest = ary[theRow][index]; } return lowest; } double GetHighestInCol(const double ary[][MaxCols], int theCol) { double highest = 0; for(int index = 0; index < MaxRows; index++) { if(ary[index][theCol] > highest) highest = ary[index][theCol]; } return highest; } double GetLowestInCol(const double ary[][MaxCols], int theCol) { double lowest = 0; for(int index = 0; index < MaxRows; index++) { if(ary[index][theCol] < lowest) lowest = ary[index][theCol]; } return lowest; } double GetHighest(const double ary[][MaxCols], int& theRow, int& theCol) { theRow = 0; theCol = 0; double highest = ary[theRow][theCol]; for(int index = 0; index < MaxRows; index++) { for(int index1 = 0; index1 < MaxCols; index1++) { double highest = 0; if(ary[index1][theCol] > highest) { highest = ary[index][index1]; theRow = index; theCol = index1; } } } return highest; } double Getlowest(const double ary[][MaxCols], int& theRow, int& theCol) { theRow = 0; theCol = 0; double lowest = ary[theRow][theCol]; for(int index = 0; index < MaxRows; index++) { for(int index1 = 0; index1 < MaxCols; index1++) { double lowest = 0; if(ary[index1][theCol] < lowest) { lowest = ary[index][index1]; theRow = index; theCol = index1; } } } return lowest; } . 1>------ Build started: Project: teddy lab 10, Configuration: Debug Win32 ------ 1>Compiling... 1>lab 10.cpp 1>c:\users\owner\documents\visual studio 2008\projects\teddy lab 10\teddy lab 10\ lab 10.cpp(46) : warning C4700: uninitialized local variable 'theRow' used 1>c:\users\owner\documents\visual studio 2008\projects\teddy lab 10\teddy lab 10\ lab 10.cpp(62) : warning C4700: uninitialized local variable 'theCol' used 1>Linking... 1> lab 10.obj : error LNK2028: unresolved token (0A0002E0) "double __cdecl GetLowest(double const (* const)[10],int &,int &)" (?GetLowest@@$$FYANQAY09$$CBNAAH1@Z) referenced in function "int __cdecl main(void)" (?main@@$$HYAHXZ) 1> lab 10.obj : error LNK2019: unresolved external symbol "double __cdecl GetLowest(double const (* const)[10],int &,int &)" (?GetLowest@@$$FYANQAY09$$CBNAAH1@Z) referenced in function "int __cdecl main(void)" (?main@@$$HYAHXZ) 1>C:\Users\owner\Documents\Visual Studio 2008\Projects\ lab 10\Debug\ lab 10.exe : fatal error LNK1120: 2 unresolved externals 1>Build log was saved at "file://c:\Users\owner\Documents\Visual Studio 2008\Projects\ lab 10\teddy lab 10\Debug\BuildLog.htm" 1>teddy lab 10 - 3 error(s), 2 warning(s) ========== Build: 0 succeeded, 1 failed, 0 up-to-date, 0 skipped ==========

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  • How does this code break the Law of Demeter?

    - by Dave Jarvis
    The following code breaks the Law of Demeter: public class Student extends Person { private Grades grades; public Student() { } /** Must never return null; throw an appropriately named exception, instead. */ private synchronized Grades getGrades() throws GradesException { if( this.grades == null ) { this.grades = createGrades(); } return this.grades; } /** Create a new instance of grades for this student. */ protected Grades createGrades() throws GradesException { // Reads the grades from the database, if needed. // return new Grades(); } /** Answers if this student was graded by a teacher with the given name. */ public boolean isTeacher( int year, String name ) throws GradesException, TeacherException { // The method only knows about Teacher instances. // return getTeacher( year ).nameEquals( name ); } private Grades getGradesForYear( int year ) throws GradesException { // The method only knows about Grades instances. // return getGrades().getForYear( year ); } private Teacher getTeacher( int year ) throws GradesException, TeacherException { // This method knows about Grades and Teacher instances. A mistake? // return getGradesForYear( year ).getTeacher(); } } public class Teacher extends Person { public Teacher() { } /** * This method will take into consideration first name, * last name, middle initial, case sensitivity, and * eventually it could answer true to wild cards and * regular expressions. */ public boolean nameEquals( String name ) { return getName().equalsIgnoreCase( name ); } /** Never returns null. */ private synchronized String getName() { if( this.name == null ) { this.name == ""; } return this.name; } } Questions How is the LoD broken? Where is the code breaking the LoD? How should the code be written to uphold the LoD?

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  • Proving that the distance values extracted in Dijkstra's algorithm is non-decreasing?

    - by Gail
    I'm reviewing my old algorithms notes and have come across this proof. It was from an assignment I had and I got it correct, but I feel that the proof certainly lacks. The question is to prove that the distance values taken from the priority queue in Dijkstra's algorithm is a non-decreasing sequence. My proof goes as follows: Proof by contradiction. Fist, assume that we pull a vertex from Q with d-value 'i'. Next time, we pull a vertex with d-value 'j'. When we pulled i, we have finalised our d-value and computed the shortest-path from the start vertex, s, to i. Since we have positive edge weights, it is impossible for our d-values to shrink as we add vertices to our path. If after pulling i from Q, we pull j with a smaller d-value, we may not have a shortest path to i, since we may be able to reach i through j. However, we have already computed the shortest path to i. We did not check a possible path. We no longer have a guaranteed path. Contradiction.

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  • pointers to functions

    - by DevAno1
    I have two basic Cpp tasks, but still I have problems with them. First is to write functions mul1,div1,sub1,sum1, taking ints as arguments and returning ints. Then I need to create pointers ptrFun1 and ptrFun2 to functions mul1 and sum1, and print results of using them. Problem starts with defining those pointers. I thought I was doing it right, but devcpp gives me errors in compilation. #include <iostream> using namespace std; int mul1(int a,int b) { return a * b; } int div1(int a,int b) { return a / b; } int sum1(int a,int b) { return a + b; } int sub1(int a,int b) { return a - b; } int main() { int a=1; int b=5; cout << mul1(a,b) << endl; cout << div1(a,b) << endl; cout << sum1(a,b) << endl; cout << sub1(a,b) << endl; int *funPtr1(int, int); int *funPtr2(int, int); funPtr1 = sum1; funPtr2 = mul1; cout << funPtr1(a,b) << endl; cout << funPtr2(a,b) << endl; system("PAUSE"); return 0; } 38 assignment of function int* funPtr1(int, int)' 38 cannot convertint ()(int, int)' to `int*()(int, int)' in assignment Task 2 is to create array of pointers to those functions named tabFunPtr. How to do that ?

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  • Fread binary file dynamic size string [C]

    - by Blackbinary
    I've been working on this assignment, where I need to read in "records" and write them to a file, and then have the ability to read/find them later. On each run of the program, the user can decide to write a new record, or read an old record (either by Name or #) The file is binary, here is its definition: typedef struct{ char * name; char * address; short addressLength, nameLength; int phoneNumber; }employeeRecord; employeeRecord record; The way the program works, it will store the structure, then the name, then the address. Name and address are dynamically allocated, which is why it is necessary to read the structure first to find the size of the name and address, allocate memory for them, then read them into that memory. For debugging purposes I have two programs at the moment. I have my file writing program, and file reading. My actual problem is this, when I read a file I have written, i read in the structure, print out the phone # to make sure it works (which works fine), and then fread the name (now being able to use record.nameLength which reports the proper value too). Fread however, does not return a usable name, it returns blank. I see two problems, either I haven't written the name to the file correctly, or I haven't read it in correctly. Here is how i write to the file: where fp is the file pointer. record.name is a proper value, so is record.nameLength. Also i am writing the name including the null terminator. (e.g. 'Jack\0') fwrite(&record,sizeof record,1,fp); fwrite(record.name,sizeof(char),record.nameLength,fp); fwrite(record.address,sizeof(char),record.addressLength,fp); And i then close the file. here is how i read the file: fp = fopen("employeeRecord","r"); fread(&record,sizeof record,1,fp); printf("Number: %d\n",record.phoneNumber); char *nameString = malloc(sizeof(char)*record.nameLength); printf("\nName Length: %d",record.nameLength); fread(nameString,sizeof(char),record.nameLength,fp); printf("\nName: %s",nameString); Notice there is some debug stuff in there (name length and number, both of which are correct). So i know the file opened properly, and I can use the name length fine. Why then is my output blank, or a newline, or something like that? (The output is just Name: with nothing after it, and program finishes just fine) Thanks for the help.

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  • big O notation algorithm

    - by niggersak
    Use big-O notation to classify the traditional grade school algorithms for addition and multiplication. That is, if asked to add two numbers each having N digits, how many individual additions must be performed? If asked to multiply two N-digit numbers, how many individual multiplications are required? . Suppose f is a function that returns the result of reversing the string of symbols given as its input, and g is a function that returns the concatenation of the two strings given as its input. If x is the string hrwa, what is returned by g(f(x),x)? Explain your answer - don't just provide the result!

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